The

*digital root* of a non-negative integer is the single digit value we get if we find the sum of its digits and keep doing this on the result until we end up with a single digit answer.

Examples:

The digital root of 25 is 2+5 = 7.

For 123, 1+2+3 = 6, so the digital root of 123 is 6.

For 578, 5+7+8 = 20. This isn't a single digit yet, so apply the process again to this number (20): 2+0 = 2. This is now a single digit, so the digital root of 578 is 2.

It can be shown that every non-negative integer has a digital root (i.e. for any non-negative integer, this iterative process will always eventually terminate). We denote the digital root of a non-negative integer a as dr(a).

One fact about digital roots is that dr(a+b) = dr(dr(a) + dr(b)) for all non-negative integers a and b. Note that by induction it follows that

dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (*)

for any positive integer k.

For example, we found above that dr(123) = 6 and dr(578) = 2. So if we wanted to find dr(123 + 578) = dr(701), we have

dr(701) = dr(dr(123) + dr(578)) = dr(6 + 2) = dr(8) = 8,

which is of course indeed correct, since 7+0+1 = 8.

Now, another fact is that a positive integer is divisible by 9 if and only if its digital root is 9. Also note that by construction, if a non-negative integer has digit sum S, then dr(a) = dr(S).

Now, assume we take a bunch of numbers a1, a2, ..., ak, that are as described by you (consisting precisely of the digits from 0 to 9, with each appearing exactly once). Let Sj be the digit sum of aj for each j = 1,2,...,k. Then

dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (by (*))

= dr(dr(S1) + dr(S2) + ... + dr(Sk)) (remembering dr(a) = dr(S), where S is the digit sum of a)

= dr(S1 + S2 + ... + Sk) (using (*) again).

Now, S1 + S2 + ... + Sk is just 0+1+2+...+9, because we are summing up digits from each of our numbers, but the numbers are constructed using the digits 0-9. But 0+1+2+...+9 is divisble by 9, so its digital root is 9, and thus we get

dr(a1 + a2 + ... + ak) = dr(S1 + S2 + ... + Sk) = 9,

so a1 + a2 + ... + ak is divisible by 9, as conjectured.

Note that we could replace your choice of digits to use (0-9) with any bunch of digits or numbers whose digit sum is divisible by 9 and we'd get the same result (because dr(a1 + ... + ak) will always equal dr(S1 + ... + Sk)).

Digital roots:

https://en.wikipedia.org/wiki/Digital_root .

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